# Revisit “Inequality between Frobenius and nuclear norm”

I am reading the following question:

Inequality between Frobenius and nuclear norm

I know $$\|X\|_* = \sum_{i=1}^r \sigma_i(X)$$and $$\|X\|_F = \bigg(\sum_{i=1}^r \sigma^2_i(X)\bigg)^{1/2}$$

I try to prove it by Cauchy-Schwarz inequality; however, I cannot see how to use C-S inequality to prove this. Could anyone please give me a small hint? Thanks in advanced.

Cauchy—Schwarz tells you that $$\sum_{i=1}^r a_i b_i \leq \left(\sum_{i=1}^r a_i^2\right)^{1/2}\left(\sum_{i=1}^r b_i^2\right)^{1/2}$$ Apply it with $$a_i = 1$$, $$b_i = \sigma_i(X)$$ (for $$1\leq i\leq r$$).